Top 10 Current Divider Rule MCQ Problems

Basic Electrical Engineering MCQ Questions on Current divider rule.

SI unit of current is

SI unit of current is:

1. volts
2. ohms
3. watts
4. amps

Current always divide in

Current always divides in:

1. Series
2. Parallel

Current divider rule is applicable on

Current divider rule is applicable on:

1. Series circuits
2. Parallel circuits

For two parallel resistors having same resistance the current will

For two parallel resistors having same resistance the current will:

1. Remain same
2. Divide

Which of the following is correct formula for current divider

Which of the following is correct formula for current divider:

1. Ix = (Rx/Req) * It
2. Ix = (Req/Rx) * It
3. Both of these
4. None of these

Correct answer: 2. Ix = (Req/Rx) * It

Current Divider Rule MCQ#6

For current divider circuit give below, if R1 = R2 then:

1. I1 > I2
2. I1 < I2
3. I1 = I2
4. Insufficient data

Correct answer: 3. I1 = I2

Current Divider Rule MCQ#7

For the same circuit given in problem 6, if R1 > 10R2, for any value of source current, the majority of current will flow through:

1. R1
2. R2

Current Divider Rule MCQ#8

Current divider rule is never applied to series circuits because:

1. It requires application of very complex formulae
2. Current never divides in series

Correct answer: 2. Current never divides in series

Current Divider Rule MCQ#9

For current divider circuit in problem 6, If Is = 2 A and R1 = 3 Ω and R2 = 2 Ω and I1 and I2 are values of currents flowing through R1 and R2, then:

1. I1 = 1.2 Ω, I2 = 0.8 Ω
2. I1 = 1.5 Ω, I2 = 0.5 Ω
3. I1 = 1.3 Ω, I2 = 0.7 Ω
4. I1 = 0.8 Ω, I2 = 1.2 Ω

Correct answer: 4. I1 = 0.8 Ω, I2 = 1.2 Ω

Current Divider MCQ#10

For circuit given in problem 6, if another resistor is added in series to R2, the current flowing through R1 will:

1. Increase
2. Decrease
3. Remain same

By increasing resistance there will be reduction of current in R2. To understand this Consider initially current source is of 5A and resistor R1 is of 2 ohms, while R2 is also 2 ohms.

Now current across R1 is: 2.5 A; (I1 = (Req/R1) * It = (1/2) * 5 A = 2.5 A

Now suppose another resistor (having resistance 1 ohms is added in series to R2. Now current across R1:

I1 = (Req/R1) * It = (1.2/2) * 5 A = 3 A